Assume that a female fly that has disrupted wings ( dsr ) and a speck body ( sp

Question

Assume that a female fly that has disrupted wings (dsr) and a speck body (sp) is mated to a male that has cinnabar eyes (cn).

Phenotypically wild-type F1 female progeny were mated to males that had speck bodies, disrupted wings and cinnabar eyes, and the following progeny were observed.

From this information, determine which gene is in the middle.

Part B

With respect to the three genes mentioned in the problem, what are the genotypes of the parents used in making the phenotypically wild-type F1 heterozygote?

Part C

What is the map distance between sp and dsr?

Express your answer to the nearest whole number.

Part D

What is the map distance between dsr and cn?

Express your answer to the nearest whole number.

Phenotype Number of offspring wild-type 112 disrupted wings 52 speck body 22 cinnabar eyes 235 disrupted wings, speck body 241 disrupted wings, cinnabar eyes 25 speck body, cinnabar eyes 46 disrupted wings, speck body, withered eyes 117

Explanation / Answer

PART A-

The maximum number of offspring’s represents PARENTAL GAMETES which would be –

   cn sp+ dsr+ / cn+ sp dsr

Now the least one are for DOUBLE CROSSOVER and as per the table the gametes would be-

cn sp+ dsr / cn+ sp dsr+

Now there is a point in Double crossover that the middle allele location is only reversed. So if we compare the above mentioned gametes, we will see that “dsr” is the one having its location reversed. That means opposite in parental and double crossover gametes.

So the sequence of gametes is – cn dsr sp for wild type.

PART C-

To find the distance between two genes we will take frequency of SINGLE CROSS over between those two genes and frequency of DOUBLE CROSS over. For sp and dsr, we have SCO= [speck body, cinnabar eyes] & [disrupted wings] & DCO= [speck body] + [disrupted wings, cinnabar eyes]

Map distance = [52 + 46 + 22 + 25 / 850] * 100 = 17.05 cM.

Here 850 is total number of offspring.

PART D-

For dsr and cn-

SCO= [wild-type] + [disrupted wings, speck body, withered eyes]

DCO = [speck body] + [disrupted wings, cinnabar eyes]

Map distance = [112+117+22+25 / 850] * 100 = 32.4 cM.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.